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tan(15°) =___________________(a) 2 + \(\sqrt{3}\)(b) 2 – \(\sqrt{3}\)(c) 1 + \(\sqrt{3}\)(d) \(\sqrt{3}\) – 1I got this question at a job interview.This interesting question is from Trigonometric Functions of Sum and Difference of Two Angles-1 in chapter Trigonometric Functions of Mathematics – Class 11

Answer»

The CORRECT answer is (b) 2 – \(\sqrt{3}\)

The best explanation: We KNOW, TAN (x -y) = (tan x – tan y)/(1+ tan x tan y)

tan (45°-30°) = (tan 45° – tan 30°)/(1+ tan 45° tan 30°)

tan 75° = (1- 1/\(\sqrt{3}\))/ (1+ 1/\(\sqrt{3}\)) = (\(\sqrt{3}\) – 1)/ (\(\sqrt{3}\) + 1) = 2-\(\sqrt{3}\).


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